We have $\operatorname{tr}(A+B) = \operatorname{tr}(A) + \operatorname{tr}(B)$ and $\det(AB) = \det(A) \det(B)$. Are there any analogous identities for the other coefficients of the characteristic polynomial?

1 Answer
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The "correct" generalization of the multiplicativity of the determinant, in my opinion, is the functoriality of the exterior powers. Namely, if $V$ is a vector space of dimension $n$ and $T : V \to V$ is a linear transformation, then the linear transformation $\Lambda^n(T) : \Lambda^n(V) \to \Lambda^n(V)$ is multiplication by $\det(T)$, and the functoriality of $\Lambda^n$ in this case says precisely that determinants are multiplicative.

The connection to the coefficients of the characteristic polynomial is that the $k^{th}$ coefficient is given by $(-1)^k \text{tr}(\Lambda^k(T))$. Thus passing from exterior powers to their traces loses some information (and this is bad for identities because the exterior powers are "multiplicative" in a certain generalized sense while the trace is additive) and to restore the nice properties you need to work with the entire exterior power.

If you work with the entire exterior power, then the most general thing you can say (using functoriality) is that if $U, V, W$ are three vector spaces and $T : U \to V, S : V \to W$ are two linear transformations, then
$$\Lambda^k(S \circ T) = \Lambda^k(S) \circ \Lambda^k(T)$$

as linear operators $\Lambda^k(U) \to \Lambda^k(W)$. Picking bases of $U, V, W$ and writing down the corresponding matrices of $T, S$ together with the corresponding bases and matrices on exterior powers then gives you a rather complicated identity (which is why it's easier to use this invariant notation instead).

A special case which might be more understandable than the general case is when $U = V = W$ has dimension $n$ and $k = n-1$. In this case we recover (more or less) the multiplicativity of the adjugate. This is a pain to write out explicitly and I don't think you will learn anything from doing so, but if you want to try anyway, take $n = 3$.

From this answer, I discovered an interesting fact about the adjugate I didn't know. It's not actually multiplicative, but $B\,{\rm adj}(AB) \,B^{-1} = {\rm adj}(A) {\rm adj}(B)$, so it's "multiplicative up to conjugacy".
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TedMar 17 '12 at 6:56

@Ted: no, the adjugate is multiplicative, but the multiplicativity looks like $\text{adj}(AB) = \text{adj}(B) \text{adj}(A)$ because of some contravariance that is needed to identify the adjugate with the exterior power; see mathoverflow.net/questions/89069/… .
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Qiaochu YuanMar 17 '12 at 8:51